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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Unit 1: Ordinary Different
ial Equations - Part 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 24 "Chapter 5: Second-Order " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "Section 5.8: the bobbing cylind
er" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 9 "Copyright" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 48 "Copyright * 2001 by Addison Wesley Longman, Inc." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 302 "All righ
ts reserved.  No part of this publication may be reproduced, stored in
 a retrieval system, or transmitted, in any form or by any means, elec
tronic, mechanical, photocopying, recording, or otherwise, without the
 prior written permission of the publisher. Printed in the United Stat
es of America." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 
0 "" {TEXT -1 15 "Initializations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Introductio
n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
85 "Undamped free motion for a spring-mass system is modeled by the di
fferential equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 273 "" 0 
"" {TEXT -1 1 " " }{XPPEDIT 18 0 "m*`y''`+k*y=0" "6#/,&*&%\"mG\"\"\"%$
y''GF'F'*&%\"kGF'%\"yGF'F'\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "Such a system is called a
 " }{TEXT 265 19 "harmonic oscillator" }{TEXT -1 47 " because the solu
tion can be put into the form " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 274 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t)=a*cos(omega*t-ph
i)" "6#/-%\"yG6#%\"tG*&%\"aG\"\"\"-%$cosG6#,&*&%&omegaGF*F'F*F*%$phiG!
\"\"F*" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "omega=sqrt(k/m)" "6#/%&omegaG
-%%sqrtG6#*&%\"kG\"\"\"%\"mG!\"\"" }{TEXT -1 268 " is the angular freq
uency.  But nature provides more instances of harmonic oscillators tha
n just spring-mass systems.  In this section we examine the motion of \+
a floating cylinder which bobs under the influence of buoyancy, a syst
em governed by an equation of the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 275 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`y''`+omega^2*y=0
" "6#/,&%$y''G\"\"\"*&%&omegaG\"\"#%\"yGF&F&\"\"!" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "and which
 is therefor a harmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "Problem Statement" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "A \+
cylinder of height " }{TEXT 266 1 "H" }{TEXT -1 9 ", radius " }{TEXT 
267 1 "R" }{TEXT -1 13 ", and weight " }{TEXT 257 1 "f" }{TEXT -1 74 "
 floats with its axis vertical in a fluid whose weight per unit volume
 is " }{XPPEDIT 18 0 "rho[w]" "6#&%$rhoG6#%\"wG" }{TEXT -1 6 ".  If " 
}{TEXT 258 1 "h" }{TEXT -1 172 " is the height, at equilibrium, of the
 submerged part of the cylinder, describe the motion of the cylinder s
ubsequent to its being  pushed down into the fluid and released." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 "The foll
owing figure (Figure 5.10) shows the waterline as Plane 1 (blue), the \+
base of the cylinder as Plane 2 (red), the height " }{XPPEDIT 18 0 "H
" "6#%\"HG" }{TEXT -1 34 " (green), the depth of submersion " }
{XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 25 " (yellow) and the radius " }
{XPPEDIT 18 0 "R" "6#%\"RG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 555 "p1 := cylinderpl
ot([1,t,z],t=0..2*Pi,z=0..3, orientation =[45,77],style=patchnogrid):
\np2 := plot3d(2,x=-2..2,y=-2..2,color=blue):\np3 := plot3d(0,x=-2..2,
y=-2..2,color=red):\np4 := spacecurve([[0,0,3],[0,-1,3]],color=black,t
hickness=3):\np5 := spacecurve(\{[0,2,t,t=0..3],[[0,2,3],[0,1,3]]\},co
lor=green, thickness=3):\np6 := spacecurve([2,0,t,t=0..2],color=yellow
,thickness=3):\np7 := textplot3d(\{[0,2.3,2.75,`H`],[2.2,0,1.75,`h`],[
0,-.6,3.3,`R`]\}, font=[TIMES,BOLD,14]):\np8 := textplot3d(\{[-1.8,1,1
.5,`Plane 1`],[-1.8,1,.3,`Plane 2`]\}):\ndisplay3d([p||(1..8)]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "
" {TEXT -1 8 "Analysis" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 55 "This mechanical problem yields to Newton'
s Second Law, " }{XPPEDIT 18 0 "F=m*a" "6#/%\"FG*&%\"mG\"\"\"%\"aGF'" 
}{TEXT -1 161 ".  The secret is realizing that the weight of the cylin
der is a force balanced by the buoyant force exerted by the fluid.  An
d this buoyant force is governed by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT 256 20 "Archimedes Principle" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 53 "Buoyant Force on Cyli
nder = Weight of Fluid Displaced" }}{PARA 265 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "F[b]=w[f]" "6#/&%\"FG6#%\"bG&%\"wG6#%\"fG" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 169 "The equilibrium a
nd non-equilibrium configurations must be considered.  At equilibrium,
 where the amount of the cylinder beneath the surface of the fluid is \+
measured by " }{XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 46 ", the volume \+
of the submerged part is given by" }}{PARA 262 "" 0 "" {XPPEDIT 18 0 "
Pi*R^2*h" "6#*(%#PiG\"\"\"*$%\"RG\"\"#F%%\"hGF%" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "The weight of the fluid d
isplaced by this submerged volume is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 263 "" 0 "" {XPPEDIT 18 0 "w[f]=Pi*R^2*h*rho[w]" "6#/&%\"wG6#%
\"fG**%#PiG\"\"\"*$%\"RG\"\"#F*%\"hGF*&%$rhoG6#F%F*" }{TEXT -1 1 " " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Conseque
ntly, the buoyant force at equlibrium is given by" }}{PARA 264 "" 0 "
" {TEXT -1 1 " " }{XPPEDIT 18 0 "F[b]=Pi*R^2*h*rho[w]" "6#/&%\"FG6#%\"
bG**%#PiG\"\"\"*$%\"RG\"\"#F*%\"hGF*&%$rhoG6#%\"wGF*" }{TEXT -1 1 " " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "By Arch
imedes Principle, at equilibrium, the buoyant force " }{XPPEDIT 18 0 "
F[b] = w[b]" "6#/&%\"FG6#%\"bG&%\"wG6#F'" }{TEXT -1 10 " balances " }
{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 36 ", the weight of the cylinder
.  Hence" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 259 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "Pi*R^2*h*rho[w]=f" "6#/**%#PiG\"\"\"*$%\"RG\"
\"#F&%\"hGF&&%$rhoG6#%\"wGF&%\"fG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 24 "This key equality links " }{XPPEDIT 18 0 
"f" "6#%\"fG" }{TEXT -1 35 ", the weight of the cylinder, with " }
{XPPEDIT 18 0 "h" "6#%\"hG" }{TEXT -1 54 ", the amount of the cylinder
 submerged at equilibrium." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 132 "Next, consider the non-equilibrium configuration
.  For this, paint a line around the cylinder at the equilibrium \"wat
er\" line.  Let " }{XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG" }{TEXT -1 
114 " denote the amount by which this line on the cylinder is displace
d above or below the surface of the fluid.  Take " }{XPPEDIT 18 0 "y(t
)" "6#-%\"yG6#%\"tG" }{TEXT -1 197 " positive if the cylinder rises (t
he black ring is visible) and negative if the cylinder descends (only \+
the fish see the black ring).  The following figure, Figure 5.11, depi
cts this configuration." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 799 "p1 := cylinderplot([1,t,z],t=0..2*
Pi,z=0..3,orientation =[45,77],style=patchnogrid):\np2 := plot3d(1.5,x
=-2..2,y=-2..2,color=blue):\np3 := plot3d(0,x=-2..2,y=-2..2,color=red)
:\np4 := spacecurve([[0,0,3],[0,-1,3]],color=black,thickness=3):\np5 :
= spacecurve(\{[0,2,t,t=0..3],[[0,2,3],[0,1,3]]\}, color=red, thicknes
s=3):\np6 := spacecurve(\{[2,0,t,t=0..2],[[2,0,2],[1,0,2]]\},color=nav
y, thickness=3):\np7 := textplot3d(\{[0,2.3,2.75,`H`],[2.2,0,1.4,`h`],
[0,-.6,3.3,`R`]\}, font=[TIMES,BOLD,14]):\np8 := spacecurve([1.05*cos(
t),1.05*sin(t),2,t=0..2*Pi],color=black, thickness=3):\np9 := spacecur
ve([.8,.6,t,t=1.5..2],color=black,thickness=3):\np10 := spacecurve([2,
2,t,t=0..1.5],color=black,thickness=3):\np11 := textplot3d(\{[.27,1,1.
75,`y(t)>0`],[.2,1,.75,`h-y(t)`]\}, font=[TIMES,BOLD,14]):\ndisplay3d(
[p||(1..11)]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Now, " }{XPPEDIT 18 0 "F[b](t)" "6#
-&%\"FG6#%\"bG6#%\"tG" }{TEXT -1 81 ", the time-dependent buoyant forc
e on the cylinder is proportional to the length " }{XPPEDIT 18 0 "h-y(
t)" "6#,&%\"hG\"\"\"-%\"yG6#%\"tG!\"\"" }{TEXT -1 8 ".  Thus," }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {XPPEDIT 18 0 "F[b](
t)=Pi*R^2*[h-y(t)]*rho[w]" "6#/-&%\"FG6#%\"bG6#%\"tG**%#PiG\"\"\"*$%\"
RG\"\"#F-7#,&%\"hGF--%\"yG6#F*!\"\"F-&%$rhoG6#%\"wGF-" }{TEXT -1 1 " \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "We ar
e ready to invoke " }{XPPEDIT 18 0 "F=m*a" "6#/%\"FG*&%\"mG\"\"\"%\"aG
F'" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "m*`y''`*``(t)" "6#*(%\"mG\"\"\"%
$y''GF%-%!G6#%\"tGF%" }{TEXT -1 29 ", Newton's Second Law, where " }
{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 39 ", the mass of the cylinder i
s given by " }{XPPEDIT 18 0 "m=f/g" "6#/%\"mG*&%\"fG\"\"\"%\"gG!\"\"" 
}{TEXT -1 7 ", with " }{XPPEDIT 18 0 "g" "6#%\"gG" }{TEXT -1 82 " bein
g the acceleration of gravity.  Thus, summing forces on the cylinder, \+
we have" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 267 "" 0 "" {XPPEDIT 
18 0 "-f+F[b](t)" "6#,&%\"fG!\"\"-&%\"FG6#%\"bG6#%\"tG\"\"\"" }{TEXT 
-1 3 " = " }{XPPEDIT 18 0 "f/g" "6#*&%\"fG\"\"\"%\"gG!\"\"" }{TEXT -1 
2 "  " }{XPPEDIT 18 0 "`y''`*``(t)" "6#*&%$y''G\"\"\"-%!G6#%\"tGF%" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 13 "which becomes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
268 "" 0 "" {XPPEDIT 18 0 "-f+Pi*R^2*[h-y(t)]*rho[w]" "6#,&%\"fG!\"\"*
*%#PiG\"\"\"*$%\"RG\"\"#F(7#,&%\"hGF(-%\"yG6#%\"tGF%F(&%$rhoG6#%\"wGF(
F(" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "f/g" "6#*&%\"fG\"\"\"%\"gG!\"\"
" }{TEXT -1 1 " " }{XPPEDIT 18 0 "`y''`*``(t)" "6#*&%$y''G\"\"\"-%!G6#
%\"tGF%" }{TEXT -1 1 " " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 53 "But from Archimedes Law at equilibri
um we found that " }{XPPEDIT 18 0 "f=Pi*R^2*rho[w]*h" "6#/%\"fG**%#PiG
\"\"\"*$%\"RG\"\"#F'&%$rhoG6#%\"wGF'%\"hGF'" }{TEXT -1 64 ".  Making t
his substitution, and expanding the brackets, we find" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {XPPEDIT 18 0 "-Pi*R^2*rho[w]*h" 
"6#,$**%#PiG\"\"\"*$%\"RG\"\"#F&&%$rhoG6#%\"wGF&%\"hGF&!\"\"" }{TEXT 
-1 3 " + " }{XPPEDIT 18 0 "Pi*R^2*rho[w]*h" "6#**%#PiG\"\"\"*$%\"RG\"
\"#F%&%$rhoG6#%\"wGF%%\"hGF%" }{TEXT -1 3 " - " }{XPPEDIT 18 0 "Pi*R^2
*rho[w]*y(t)" "6#**%#PiG\"\"\"*$%\"RG\"\"#F%&%$rhoG6#%\"wGF%-%\"yG6#%
\"tGF%" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "f/g" "6#*&%\"fG\"\"\"%\"gG!
\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "`y''`*``(t)" "6#*&%$y''G\"\"\"-%
!G6#%\"tGF%" }{TEXT -1 1 " " }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "which collapses to" }}{PARA 
270 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "-Pi*R^2*rho[w]*y(t)=f/g" "6
#/,$**%#PiG\"\"\"*$%\"RG\"\"#F'&%$rhoG6#%\"wGF'-%\"yG6#%\"tGF'!\"\"*&%
\"fGF'%\"gGF3" }{TEXT -1 1 " " }{XPPEDIT 18 0 "`y''`*``(t)" "6#*&%$y''
G\"\"\"-%!G6#%\"tGF%" }{TEXT -1 1 " " }{TEXT -1 1 " " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "Hence, we have the equa
tion" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 276 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`y''`" "6#%$y''G" }
{TEXT -1 3 " + " }{XPPEDIT 18 0 "Pi*R^2*rho[w]*g/f" "6#*,%#PiG\"\"\"*$
%\"RG\"\"#F%&%$rhoG6#%\"wGF%%\"gGF%%\"fG!\"\"" }{TEXT -1 1 " " }{TEXT 
259 1 "y" }{TEXT -1 5 " = 0 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 19 " which we write as " }{XPPEDIT 18 0 "`y''`+omeg
a^2" "6#,&%$y''G\"\"\"*$%&omegaG\"\"#F%" }{TEXT 260 1 "y" }{TEXT -1 
16 " = 0 by defining" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 271 "" 
0 "" {XPPEDIT 18 0 "omega=R*sqrt(Pi*rho[w]*g/f)" "6#/%&omegaG*&%\"RG\"
\"\"-%%sqrtG6#**%#PiGF'&%$rhoG6#%\"wGF'%\"gGF'%\"fG!\"\"F'" }{TEXT -1 
1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "
We therefore have simple harmonic motion governed by the initial value
 problem for undamped free motion" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 260 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "`y''`+omega^2" "6#,&%
$y''G\"\"\"*$%&omegaG\"\"#F%" }{TEXT -1 5 "y = 0" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 10 "   y(0) = " }{XPPEDIT 
18 0 "y[0]" "6#&%\"yG6#\"\"!" }{TEXT -1 1 " " }}{PARA 277 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "`y'`*``(0)=v[0]" "6#/*&%#y'G\"\"\"-%!G6
#\"\"!F&&%\"vG6#F*" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "
" {TEXT -1 8 "Solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 46 "The solution to this initial value proble
m is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 278 "" 0 "" {TEXT -1 1 
" " }{XPPEDIT 18 0 "y(t)=y[0]*cos(omega*t)+v[0]/omega" "6#/-%\"yG6#%\"
tG,&*&&F%6#\"\"!\"\"\"-%$cosG6#*&%&omegaGF-F'F-F-F-*&&%\"vG6#F,F-F2!\"
\"F-" }{TEXT -1 1 " " }{XPPEDIT 18 0 "sin(omega*t)" "6#-%$sinG6#*&%&om
egaG\"\"\"%\"tGF(" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 77 "which says the cylinder bobs with periodi
c motion whose angular frequency is " }{XPPEDIT 18 0 "omega" "6#%&omeg
aG" }{TEXT -1 21 " and whose period is " }{XPPEDIT 18 0 "T=2*Pi/omega
" "6#/%\"TG*(\"\"#\"\"\"%#PiGF'%&omegaG!\"\"" }{TEXT -1 38 ".  The sol
ution can also be written as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
279 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t)=sqrt(``(v[0])^2+omega^
2)/omega" "6#/-%\"yG6#%\"tG*&-%%sqrtG6#,&*$-%!G6#&%\"vG6#\"\"!\"\"#\"
\"\"*$%&omegaGF5F6F6F8!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "cos(omega
*t-phi)" "6#-%$cosG6#,&*&%&omegaG\"\"\"%\"tGF)F)%$phiG!\"\"" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 
"where " }{XPPEDIT 18 0 "phi" "6#%$phiG" }{TEXT -1 18 " is determined \+
by " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 280 "" 0 "" {TEXT -1 1 " \+
" }{XPPEDIT 18 0 "sin(phi)=v[0]/sqrt(``(v[0])^2+omega^2*``(y[0])^2)" "
6#/-%$sinG6#%$phiG*&&%\"vG6#\"\"!\"\"\"-%%sqrtG6#,&*$-%!G6#&F*6#F,\"\"
#F-*&%&omegaGF8-F46#&%\"yG6#F,F8F-!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 281 "" 
0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "cos(phi)=omega*y[0]/sqrt(``(v[0])^
2+omega^2*``(y[0])^2)" "6#/-%$cosG6#%$phiG*(%&omegaG\"\"\"&%\"yG6#\"\"
!F*-%%sqrtG6#,&*$-%!G6#&%\"vG6#F.\"\"#F**&F)F:-F56#&F,6#F.F:F*!\"\"" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "The reader is again cauti
oned not to define " }{XPPEDIT 18 0 "phi" "6#%$phiG" }{TEXT -1 4 " as \+
" }{XPPEDIT 18 0 "arctan(v[0]/omega/y[0])" "6#-%'arctanG6#*(&%\"vG6#\"
\"!\"\"\"%&omegaG!\"\"&%\"yG6#F*F-" }{TEXT -1 47 " since the range of \+
the arctangent function is " }{XPPEDIT 18 0 "``(-Pi/2,Pi/2)" "6#-%!G6$
,$*&%#PiG\"\"\"\"\"#!\"\"F+*&F(F)F*F+" }{TEXT -1 6 ", not " }}{PARA 0 
"" 0 "" {TEXT -1 1 "(" }{XPPEDIT 18 0 "-Pi,Pi" "6$,$%#PiG!\"\"F$" }
{TEXT -1 42 "].  A negative value for the displacement " }{XPPEDIT 18 
0 "y[0]" "6#&%\"yG6#\"\"!" }{TEXT -1 64 " results in a second-quadrant
 phase angle, while a negative for " }{XPPEDIT 18 0 "v[0]" "6#&%\"vG6#
\"\"!" }{TEXT -1 147 ", the initial velocity, results in a third-quadr
ant angle.  These angles would not be correctly obtained by a naive us
e of the arctangent function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 73 "To verify these calculations in Maple, en
ter the differential equation as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "q := diff(y(t),t,t) + omega^
2*y(t) = 0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 15 "and obtain the " }{TEXT 261 6 "dsolve" }
{TEXT -1 9 " solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "Y := rhs(dsolve(\{q, y(0) = y0, D(y
)(0)=v0\}, y(t)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "which says the cylinder bobs with \+
periodic motion whose angular frequency is " }{XPPEDIT 18 0 "omega" "6
#%&omegaG" }{TEXT -1 25 " and whose period is T = " }{XPPEDIT 18 0 "2*
Pi/omega" "6#*(\"\"#\"\"\"%#PiGF%%&omegaG!\"\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 186 "Addition
ally, we can write the solution in terms of a single trig function.  T
his makes it easier to determine the amplitude of motion.  The templat
e for the single-trig-term solution is" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "YY := A*cos(omega*t - p
hi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 52 "To get Maple to make the match to this template, \+
try" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "q1 := solve(identity(Y = expand(YY),t),\{A,phi\});" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 20 "Maple has expressed " }{XPPEDIT 18 0 "phi" "6#%$phiG" }
{TEXT -1 83 ", the phase angle, in terms of the two-argument arctangen
t function.  The function " }{XPPEDIT 18 0 "arctan(y/x)" "6#-%'arctanG
6#*&%\"yG\"\"\"%\"xG!\"\"" }{TEXT -1 32 " returns an angle in the rang
e (" }{XPPEDIT 18 0 "-Pi/2,Pi/2" "6$,$*&%#PiG\"\"\"\"\"#!\"\"F(*&F%F&F
'F(" }{TEXT -1 68 "] which includes only the first and fourth quadrant
s.  The function " }{XPPEDIT 18 0 "arctan(y,x)" "6#-%'arctanG6$%\"yG%
\"xG" }{TEXT -1 71 ", the two-argument arctangent function, returns an
 angle in the range (" }{XPPEDIT 18 0 "-Pi,Pi" "6$,$%#PiG!\"\"F$" }
{TEXT -1 93 "] which includes all four quadrants.  Maple's usage is pr
ecise, but for symbolic work, messy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 50 "The amplitude is taken as positive, so \+
we conclude" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 27 "amplitude =  subs(q1[1],A);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Exam
ple 5.13" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 21 "A cylinder of radius " }{XPPEDIT 18 0 "R=2" "6#/%\"RG\"\"
#" }{TEXT -1 83 " inches bobs, with a period of 1/2 second, in water w
hose weight per cubic foot is " }{XPPEDIT 18 0 "rho[w]" "6#&%$rhoG6#%
\"wG" }{TEXT -1 47 " = 62.5 lbs.  Using the gravitational constant " }
{XPPEDIT 18 0 "g=32 " "6#/%\"gG\"#K" }{TEXT -1 1 " " }{XPPEDIT 18 0 "f
t/sec^2" "6#*&%#ftG\"\"\"*$%$secG\"\"#!\"\"" }{TEXT -1 8 " , find " }
{TEXT 262 1 "f" }{TEXT -1 62 ", the weight of the cylinder, and find t
he equilibrium height " }{TEXT 263 1 "h" }{TEXT -1 1 "." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 222 "This problem is rea
listic in that it uses an observable, the period, from which you can d
educe information about the nature of the cylinder.  The period can be
 estimated by careful counting in the presence of a wristwatch." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "The perio
d is " }{XPPEDIT 18 0 "T=2*Pi/omega" "6#/%\"TG*(\"\"#\"\"\"%#PiGF'%&om
egaG!\"\"" }{TEXT -1 7 ", where" }}{PARA 272 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "omega=R*sqrt(Pi*rho[w]*g/f)" "6#/%&omegaG*&%\"RG\"\"\"-
%%sqrtG6#**%#PiGF'&%$rhoG6#%\"wGF'%\"gGF'%\"fG!\"\"F'" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "Hence, we have" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 31 "omega := R*sqrt(Pi*rho[w]*g/f);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 " and" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "T
 := 2*Pi/omega;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Setting " }{XPPEDIT 18 0 "T=1/2" "6
#/%\"TG*&\"\"\"F&\"\"#!\"\"" }{TEXT -1 12 " and taking " }{XPPEDIT 18 
0 "R=2/12,rho[w]=62.5" "6$/%\"RG*&\"\"#\"\"\"\"#7!\"\"/&%$rhoG6#%\"wG$
\"$D'F)" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g=32" "6#/%\"gG\"#K" }
{TEXT -1 24 ", we obtain the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "q2 := 1/2 = subs(R = 2/
12,rho[w] = 62.5,g = 32, T);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "which we solve for the w
eight " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 9 ", getting" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "w
t := solve(q2,f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "The weight of the cylinder is 1.10
524266 pounds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 31 "To find the equilibrium height " }{TEXT 264 1 "h" }{TEXT 
-1 41 ", again use the equilibrium relationship " }{XPPEDIT 18 0 "f=Pi
*R^2*rho[w]*h" "6#/%\"fG**%#PiG\"\"\"*$%\"RG\"\"#F'&%$rhoG6#%\"wGF'%\"
hGF'" }{TEXT -1 16 ".  Consequently," }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "q3 := wt = Pi*(2/12)^2 * \+
62.5 * h;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 11 "from which " }{XPPEDIT 18 0 "h" "6#%\"hG
" }{TEXT -1 50 ", the submerged length at equilibrium, follows via" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 16 "h = solve(q3,h);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The equilibrium height of the c
ylinder is .202 feet (hence, 2.43 inches)." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "1" 
0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }